1. Use the condition that $A$ lies in the $xy$-plane

Since $A(x,y,z)$ is in the $xy$-plane, $z=0.$ So let $A=(x,y,0).$

2. Given that $A$ is equidistant from $(0,3,2)$, $(2,0,3)$ and $(0,0,1)$

Let $P=(0,3,2),\quad Q=(2,0,3),\quad R=(0,0,1).$ Then $AP=AQ=AR.$ We equate the squared distances.

(i) Equate $AP^2$ and $AR^2$

$AP^2=x^2+(y-3)^2+(0-2)^2=x^2+(y-3)^2+4,$ $AR^2=x^2+y^2+(0-1)^2=x^2+y^2+1.$ So, $x^2+(y-3)^2+4=x^2+y^2+1.$ Expanding, $y^2-6y+9+4=y^2+1,$ $-6y+13=1,$ $-6y=-12,$ $y=2.$

(ii) Equate $AQ^2$ and $AR^2$

$AQ^2=(x-2)^2+y^2+(0-3)^2=(x-2)^2+y^2+9,$ $AR^2=x^2+y^2+1.$ Thus, $(x-2)^2+y^2+9=x^2+y^2+1,$ $x^2-4x+4+9=x^2+1,$ $-4x+13=1,$ $-4x=-12,$ $x=3.$

Hence, $A=(3,2,0).$

3. Coordinates of the three vertices $A=(3,2,0),\quad B=(1,4,-1),\quad C=(2,0,-2).$

4. Check statement (S1): is $\triangle ABC$ an isosceles right triangle?

Compute the side lengths squared.

(i) $AB^2$

$AB^2=(3-1)^2+(2-4)^2+(0+1)^2=2^2+(-2)^2+1^2=4+4+1=9.$ So, $AB=3.$

(ii) $AC^2$

$AC^2=(3-2)^2+(2-0)^2+(0+2)^2=1^2+2^2+2^2=1+4+4=9.$ So, $AC=3.$

(iii) $BC^2$

$BC^2=(1-2)^2+(4-0)^2+(-1+2)^2=(-1)^2+4^2+1^2=1+16+1=18.$ So, $BC=3\sqrt{2}.$

Thus, $AB=AC=3,$ so the triangle is isosceles.

Now check for right angle: $AB^2+AC^2=9+9=18=BC^2.$ Hence, by Pythagoras theorem, the angle between $AB$ and $AC$ is $90^\circ$.

Therefore, $\triangle ABC$ is an isosceles right triangle.

So (S1) is true.

5. Check statement (S2): area of $\triangle ABC$

Since the triangle is right angled at $A$ with perpendicular sides $AB=AC=3,$ its area is $\text{Area}=\frac{1}{2}\cdot AB\cdot AC=\frac{1}{2}\cdot 3\cdot 3=\frac{9}{2}.$

But the statement says area is $\frac{9\sqrt{2}}{2},$ which is incorrect.

So (S2) is false.

6. Final conclusion

• (S1) is true
• (S2) is false

Therefore, the correct option is $\boxed{\text{C}}$ which means only (S1) is true.